IODS course project
Rong Guang
Introduction to Open Data Science - Course Project
# This is a so-called "R chunk" where you can write R code.
date()## [1] "Tue Nov 29 12:54:57 2022"Chapter 1: Start me up
1. Thoughts about the course
1.1 My feeling about the course
I am feeling very good, since I finally start using GitHub from scratch. All my colleagues are using GitHub for collaboration, but I do not even have any idea what it is. This course is very timely. Hopefully after this course, I can join their collaboration loop effectively.
1.2 What do I expect to learn
I do not have any particular expectation in the statistical knowledge I am going to learn, because anything about statistics is interesting to me. Actually, I hope in the course I could have chance of exercising GitHub collaboration in practical manner. For example, my colleagues always mention the creation of some ‘branches’ of some “root” data sets, which I still do not understand very clearly what is it.
2. reflect on my learning with the R for Health Data Science book
2.1 How did it work as a “crash course” on modern R tools and using RStudio?
Luckily, I just finished taking Prof. Kimmo’s another course “Quantitative Research Skills”, where I got chance to walk through several chapters of R for Health Data Science book. This way I suppose I could make some comments on the book retrospectively. I would say this is the best book for novice R users to start off. The author did the instruction from an industry perspective ( health science), which avoid getting too deep in math but focuses on the R as a statistical tool for solving industry problem itself. When I tried to start with R years ago, one of the biggest barrier is for every statistical feature I want to realize, R always present a multitude of pathways to realize it (for example, to compute a new variable for a data set, you can you “.[,c] <-”, or “.$c<-”, or ” %>% ” + “mutate”), other source of material/textbook always tried to feed you with all these possibilities, which makes things complicated. This book just doesn’t. For each function to execute, it only show you one approach. This is just what most non-statistics-majors want!
2.2 Which were your favorite topics?
Data wrangling, definitely (although there is no specific chapter for it, but it scatters throughout the whole book). Most descriptive, inferential and exploratory statistics are so easy to realize in R. By the merit of R, for these purposes, all I need is first, know what I am going to do (base on my stat knowledge), and then search in Google about which package(s) are for doing that, and finally install it, read the instruction&example and then do it. Very mechanical. There is never a big challenge. To me, data wrangling (more than dozens of variable) requires some real effort, and hence more interesting. And there is never an end. Today I realize a wrangling task in three lines of codes, maybe tomorrow I could come up with or stumble over a way that only requires two, or even one. This process is super exciting.
2.3 Which topics were most difficult?
Data wrangling, of course according to my story above, followed by logistic regression. For logistic regression, I have several big unsolved problem on mind. They are: a. considering there is not an alternative of linear regression’s r square for logistic regression, how do I evaluate the variance that my model could explain? b. after modeling additive effect, model modification is recommended to follow. It is not difficult to create a main effect plot to observe candidate multiplicative variables if y is continuous. But is it also applicable to a binary y ? However, I suppose this is mainly due to my lack of relevant stat knowledge, having nothing to do with R language.
3. My GitHub Repo
Please find it Here: https://github.com/rg450318262/IODS-project
4. Token
Since I have generated the token before the new task is published (That is the only way I get my machine connected to GitHub), I will not generate it again in case any overlapping problems.
This is the end of chapter 1
Chapter 2: Regression and model validation
Describe the work you have done this week and summarize your learning.
- Describe your work and results clearly.
- Assume the reader has an introductory course level understanding of writing and reading R code as well as statistical methods.
- Assume the reader has no previous knowledge of your data or the more advanced methods you are using.
date()## [1] "Tue Nov 29 12:54:57 2022"1 Preparing
1.1 read the data set
library(tidyverse)
learn <- read_csv(file = "data/learning2014.csv")1.2 Code categorical data
learn <- learn %>%
mutate(gender = gender %>%
factor() %>%
fct_recode("Female" = "F",
"Male" = "M"))1.3 explore the data set
#explore dimensions
dim(learn)## [1] 166 7The data set has 166 observations of 7 variables.
#explore structure
str(learn)## tibble [166 × 7] (S3: tbl_df/tbl/data.frame)
## $ gender : Factor w/ 2 levels "Female","Male": 1 2 1 2 2 1 2 1 2 1 ...
## $ age : num [1:166] 53 55 49 53 49 38 50 37 37 42 ...
## $ attitude: num [1:166] 3.7 3.1 2.5 3.5 3.7 3.8 3.5 2.9 3.8 2.1 ...
## $ deep : num [1:166] 3.58 2.92 3.5 3.5 3.67 ...
## $ stra : num [1:166] 3.38 2.75 3.62 3.12 3.62 ...
## $ surf : num [1:166] 2.58 3.17 2.25 2.25 2.83 ...
## $ points : num [1:166] 25 12 24 10 22 21 21 31 24 26 ...The data set has six numeric (integer type) variables and one categorical (binary) variable.
1.4 describe the variables
Under the funding of International Survey of Approaches to Learning, 183 Finnish students who took the course “Introduction to Social Statistics” during 2014 fall participated in a survey about their learning, resulting in a data set with 32 variables and 166 observations (due to missing data points, the sample size is smaller than 183). The current data set for analysis is a convenient subset of it. It includes variables about the participants’ demographic characteristics such as age and sex, as well as the final points they got for certain exam (could possibly be statistics). It also includes 4 psychological dimensions including study attitude (reflecting their motivation to the subject), deep learning score (reflecting how well their learning style fits into the deep learning type), surface learning score (reflecting how well their learning style fits into the surface learning type) and strategy learning score (reflecting how well their learning style fits into the strategic learning type).
1.4.1 describe the coninuous variable
library(tidyverse)
library(finalfit) # a package introduced in RHDS book.
#The "gg_glimpse" function could give nice descriptive
#statistics for both types of variables.
library(DT) # show table in a html-based neat view.
ff_glimpse(learn)$Continuous %>% datatable() # descriptive statistics for #categorical data shown in html-based data table view.According to their distribution shapes visualized in section 1.5 (next section), non-normally distributed variables were reported as median and Q1-Q3; roughly normal variables will be reported as mean±sd.
The age of the participants was between 17 and 55 years old (median: 22; Q1-Q3:21,27 years old). Their exam points were 22.7±5.9. Their deep learning scores were 3.7±0.6. Their surface learning scores were 2.8±0.5. And their strategic learning scores were 3.1±0.8.
1.4.2 describe the categorical variable
ff_glimpse(learn)$Categorical %>% datatable() # descriptive statistics for categorical data shown in html-based data table view.Among the 166 participants, 110 (66%) were female and 56 (34%) were male. According to a 2021 statistics, Finnish universities had a male:female ratio of 1:1.2, indicating female in our sample is over-represented.
1.5 visualize the data set
library(GGally)
library(ggplot2)
library(tidyverse)
#create a self-defined function so that correlation matrix produced by ggpairs could
#show LOWESS smoothing with scatter plot.
my_fn <- function(data, mapping, method="loess", ...){ #require the input of two
#arguments: data and mapping; arguments method is set to be "Loess"
#for more information about using Loess to check function form, please
#go to (https://thestatsgeek.com/2014/09/13/checking-functional-form-in-logistic-regression-using-loess/)
p <- ggplot(data = data, mapping = mapping) + #call ggplot function
geom_point(size = 0.3, color = "coral") + #call point graph, reduce the
#size, turn color to coral, for better visualization
geom_smooth(size = 0.3, method=method, ...) # fit Loess regression
p #print the result
}
# create an plot matrix with ggpairs()
ggpairs(learn,
lower= list(combo = wrap("facethist", bins = 20),
continuous = my_fn) #call self-defined function "my_fn"
)
According to the visualization, it is found that the distribution of age is right-skewed; other numeric variables, though with slight skewness, can be roughly treated as normal distribution. All of the values of numeric variables did not show any remarkable difference between males and females. Variables “points”, “attitude” and “deep” have 1 to 3 out-liers, respectively, and age has quite a number of out-liers. By examining the raw data, no evidence of mistaken record was detected. These out-liers were thus kept for analysis. Using variable “points” as reference, variable “attitude” showed a significant linear correlation(r=0.437). Although the correlation coefficient between age and points is only -0.093, the LOESS smoothing has shown there might be a quadratic relationship between them.
2. Fitting the model
2.1 variable selection
According to the visualization in section 1.4, age (as polynomial form due to its non-linearity with the outcome) and attitude were used to fit the model that predicts exam points. Although no noticeable effect of gender was observed, it also entered the model for it being adopted as an important factor for predicting exam points in a multitude of publications.
2.2 fitting
fit1 <- learn %>% #using attitude, the polynomial age and gender to predict exam points
lm(points ~ attitude + poly(age, 2, raw =T) + gender, data = .) #ploy() is to
#include 2nd order function form, where "2" means the order
summary(fit1) # summarize the results##
## Call:
## lm(formula = points ~ attitude + poly(age, 2, raw = T) + gender,
## data = .)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.1258 -3.1673 0.5261 3.6243 10.3486
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -4.246278 5.618253 -0.756 0.450873
## attitude 3.774918 0.576603 6.547 7.5e-10 ***
## poly(age, 2, raw = T)1 1.094988 0.345856 3.166 0.001849 **
## poly(age, 2, raw = T)2 -0.017766 0.005188 -3.424 0.000782 ***
## genderMale -0.615038 0.894211 -0.688 0.492569
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.148 on 161 degrees of freedom
## Multiple R-squared: 0.256, Adjusted R-squared: 0.2375
## F-statistic: 13.85 on 4 and 161 DF, p-value: 9.947e-10The results showed that except for gender, other variables all had significant predicting effect (all p<0.01). Besides, F-statistics (p < 0.01) had rejected the null that the response variable cannot be represented as a function of any of the predictor variables, indicating the model is valid. Adjusted R-squared showed that the model explained 23.75% of the variance of exam points. However, in the next step I further reduced the model complexity by removing insignificant variable base on the rule of parsimony.
2.3 removing insignifiant predictor
The model was fit again by removing gender.
fit2 <- learn %>%
lm(points ~ attitude + poly(age, 2, raw =T), data = .)#ploy() is to
#include 2nd order function form, where "2" means the order
summary(fit2)##
## Call:
## lm(formula = points ~ attitude + poly(age, 2, raw = T), data = .)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.904 -3.290 0.293 3.594 10.342
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -3.651986 5.542377 -0.659 0.510882
## attitude 3.656205 0.549269 6.656 4.13e-10 ***
## poly(age, 2, raw = T)1 1.068947 0.343218 3.114 0.002180 **
## poly(age, 2, raw = T)2 -0.017435 0.005158 -3.380 0.000907 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.139 on 162 degrees of freedom
## Multiple R-squared: 0.2538, Adjusted R-squared: 0.24
## F-statistic: 18.36 on 3 and 162 DF, p-value: 2.634e-10The results showed that all variables had significant predicting effect (all p<0.01). Besides, F-statistics (p < 0.01) had rejected the null that the response variable can not be represented as a function of any of the predictor variables, indicating the model is valid. Adjusted R-squared showed that the model explained 24% of the variance of exam points, which slightly outperformed the previous model. I took this model as the final model for model diagnostics.
In the final model, variable “attitude” has a coefficient of 3.65, indicating for 1 unit of attitude increase, the exam points is expected to increase 3.65, after controlling for other factors. The first order term of age has an coefficient estimate of 1.06, indicating that, overall, for every 1 unit increase of age, the exam points is expected to increase 1.06, after controlling for other factors. For the second order term of age, an estimated coefficient of -0.017 indicated for different value ranges of age, the effect on exam points might be significantly different. This auto-interaction might lead to -0.017 decrease in exam points across these ranges, after controlling for the other factors.
The practical explanation for these coefficient might be a. attitude reflects the motivation of study and higher motivation will lead to better exam performance; b. statistics requires quite a bit of domain knowledge (economics, health, psychology..), logic reasoning and math foundations. Older students might have advantage in these aspects. c. However, this advantage will see a ceiling effect at around 30 years old (according to the graph above), and due to the aging and family burden, students over 30 years old might start to become less and less competitive in stat learning over time.
3. Model diagnostic
In model diagnostic, some of the assumptions (linearity and normality) of linear regression were checked. Besides, observations with high influence will be examined in this section.
3.1 diagnostic plots
par(cex = 0.5,fig=c(0,0.5,0.5,1)) #set the coordinate of picture 1
plot(fit2, which = 1) #plot diagnostic picture 1
par(cex = 0.5,fig=c(0.5,1,0.5,1), new=TRUE)#set the coordinate of picture 2
plot(fit2, which = 2) #plot diagnostic picture 2
par(cex = 0.5, fig=c(0,1,0,0.5), new=TRUE)#set the coordinate of picture 3
plot(fit2, which = 5)#plot diagnostic picture 3
Residuals vs fitted plot (upper left) showed the data points are randomly scattered around the dotted line of y = 0, and the fitted line (red) is roughly horizontal without distinct patterns or trends, indicating a linear relationship. The linearity assumption of linear regression is met.
The QQ plot (upper right) showed most of the points plotted on the graph lies on the dashed straight line, except for the lower and upper ends, where some points deviated from the line, indicating the distribution might be slightly skewed. Considering the fact that in large sample size the assumption of linearity is almost never perfectly met, I see the assumption of normality as being approximately met.
3.2 other linear model assumptions
The assumption of independence requires no relation between the different observations. I do not have information of how this study was designed, hence not being able to make any conclusion. However, I could imagine how hard it took to meet it here, since including students taking courses from different lecturers or different lecturer groups would lead to violation of it. On the other hand, if the results were from students of one same lecturer (or lecturer group), it might take several semesters to collect such a large sample or might take students from several different classes/majors in one semester, either way the assumption was violated.
Homoscedasticity is another assumption to check. However, considering it is better evaluated by fitted values against root of standardized residuals (the #3 in plot() function), which is not required to produce in the current assignment, I did not further dig into it. By looking into its rough substitute plot “residual vs fitted” (upper left, above), no obvious heteroscedasticity was detected.
3.3 influential observations
Influential observations were shown in the bottom plot, where the red dashed line indicate cook’s distance. Cook’s distance is a commonly used estimate of the influence of a data point when performing a least-squares regression analysis. It measures the effect of deleting the observation for each given observation. In the plot, points, if there is any, outside the red dashed line are believed to have high influence. The graph for current model showed no points outside the line. The plot also showed the case numbers of 3 data points with the largest cook’s distances, which are #1, #4 and #56. However, there are also other rules of thumbs for the cutoff, which are stricter. They include using an absolute value of 1, or using 4/n (n is sample size), or using 4×(the mean of the cooks distance for the whole sample). I did not further report them since it is not required in this assignment. I did it somewhere else for fun. If you are interested, please go to a r markdown file named “Supplement_Codes.html” or “Supplement_Codes.Rmd” under my “IODS-project” folder.
of cook’s distance,
where x is the index number of our sample and y is the cook’s distance score for each observation. This is to evaluate, if there’s any, the data points being tremendously influential to the coefficient estimate. Cook’s distance is a commonly used estimate of the influence of a data point when performing a least-squares regression analysis. It measures the effect of deleting the observation for each given observation. There is no consensus on the cutoff for being influential using this indicator. The rules of thumbs include using an absolute value of 1, or using 4/n (n is sample size), or using 4×(the mean of the cooks distance for the whole sample). The plot showed the case numbers of 3 data points with the largest cook’s distances, which are #1, #4 and #56. I did not further report them since it is not required in this assignment. I did it somewhere else for fun. If you are interested, please go to a r markdown file named “Supplement_Codes.html” or “Supplement_Codes.Rmd” under my “IODS-project” folder.
This is the end of chapter 2
***********************
Here we go again…
a test to see if it works
Chapter 3: Logistic regression
1 Preparing
1.1 read the data set
library(tidyverse)
alc <- read_csv(file = "data/alc.csv")1.2 check the data set
glimpse(alc)## Rows: 370
## Columns: 43
## $ school <chr> "GP", "GP", "GP", "GP", "GP", "GP", "GP", "GP", …
## $ sex <chr> "F", "F", "F", "F", "F", "M", "M", "F", "M", "M"…
## $ age <dbl> 18, 17, 15, 15, 16, 16, 16, 17, 15, 15, 15, 15, …
## $ address <chr> "U", "U", "U", "U", "U", "U", "U", "U", "U", "U"…
## $ famsize <chr> "GT3", "GT3", "LE3", "GT3", "GT3", "LE3", "LE3",…
## $ Pstatus <chr> "A", "T", "T", "T", "T", "T", "T", "A", "A", "T"…
## $ Medu <dbl> 4, 1, 1, 4, 3, 4, 2, 4, 3, 3, 4, 2, 4, 4, 2, 4, …
## $ Fedu <dbl> 4, 1, 1, 2, 3, 3, 2, 4, 2, 4, 4, 1, 4, 3, 2, 4, …
## $ Mjob <chr> "at_home", "at_home", "at_home", "health", "othe…
## $ Fjob <chr> "teacher", "other", "other", "services", "other"…
## $ reason <chr> "course", "course", "other", "home", "home", "re…
## $ guardian <chr> "mother", "father", "mother", "mother", "father"…
## $ traveltime <dbl> 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 1, …
## $ studytime <dbl> 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 3, 1, …
## $ schoolsup <chr> "yes", "no", "yes", "no", "no", "no", "no", "yes…
## $ famsup <chr> "no", "yes", "no", "yes", "yes", "yes", "no", "y…
## $ activities <chr> "no", "no", "no", "yes", "no", "yes", "no", "no"…
## $ nursery <chr> "yes", "no", "yes", "yes", "yes", "yes", "yes", …
## $ higher <chr> "yes", "yes", "yes", "yes", "yes", "yes", "yes",…
## $ internet <chr> "no", "yes", "yes", "yes", "no", "yes", "yes", "…
## $ romantic <chr> "no", "no", "no", "yes", "no", "no", "no", "no",…
## $ famrel <dbl> 4, 5, 4, 3, 4, 5, 4, 4, 4, 5, 3, 5, 4, 5, 4, 4, …
## $ freetime <dbl> 3, 3, 3, 2, 3, 4, 4, 1, 2, 5, 3, 2, 3, 4, 5, 4, …
## $ goout <dbl> 4, 3, 2, 2, 2, 2, 4, 4, 2, 1, 3, 2, 3, 3, 2, 4, …
## $ Dalc <dbl> 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, …
## $ Walc <dbl> 1, 1, 3, 1, 2, 2, 1, 1, 1, 1, 2, 1, 3, 2, 1, 2, …
## $ health <dbl> 3, 3, 3, 5, 5, 5, 3, 1, 1, 5, 2, 4, 5, 3, 3, 2, …
## $ failures <dbl> 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, …
## $ paid <chr> "no", "no", "yes", "yes", "yes", "yes", "no", "n…
## $ absences <dbl> 5, 3, 8, 1, 2, 8, 0, 4, 0, 0, 1, 2, 1, 1, 0, 5, …
## $ G1 <dbl> 2, 7, 10, 14, 8, 14, 12, 8, 16, 13, 12, 10, 13, …
## $ G2 <dbl> 8, 8, 10, 14, 12, 14, 12, 9, 17, 14, 11, 12, 14,…
## $ G3 <dbl> 8, 8, 11, 14, 12, 14, 12, 10, 18, 14, 12, 12, 13…
## $ alc_use <dbl> 1.0, 1.0, 2.5, 1.0, 1.5, 1.5, 1.0, 1.0, 1.0, 1.0…
## $ high_use <lgl> FALSE, FALSE, TRUE, FALSE, FALSE, FALSE, FALSE, …
## $ family.quality <dbl> 4, 5, 4, 3, 4, 5, 4, 4, 4, 5, 3, 5, 4, 5, 4, 4, …
## $ in.class.performance <dbl> 5, 3, 8, 1, 2, 8, 0, 4, 0, 0, 1, 2, 1, 1, 0, 5, …
## $ off.class.performance <chr> "Moderate to long study", "Moderate to long stud…
## $ social <chr> "Frequent", "Infrequent", "Infrequent", "Infrequ…
## $ probability <dbl> 0.29068345, 0.07880612, 0.15496055, 0.14278287, …
## $ prediction <lgl> FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, TRUE, …
## $ random.guess <dbl> 0.10001649, 0.97795944, 0.73361349, 0.68607945, …
## $ prediction.guess <lgl> FALSE, TRUE, TRUE, TRUE, TRUE, FALSE, FALSE, TRU…This data approach student achievement in secondary education of two Portuguese schools. The data attributes include student grades, demographic, social and school related features) and it was collected by using school reports and questionnaires. Two datasets are provided regarding the performance in two distinct subjects: Mathematics (mat) and Portuguese language (por).
2 Hypothesis
2.1 introduction
Despite the health risk and public harm associated with heavy drinking, alcohol is the most commonly used substance in developed countries (Flor and Gakidou 2020). A large scale longitudinal study has identified Finland as the only Nordic country whose alcohol-attributable harms has increased (Room et al. 2013). Given that alcohol consumption typically starts in late adolescence or early adulthood (Lees et al. 2020), measures to detect alcohol misuse among young people, especially college students, should be a top public health priority. Identifying a comprehensive set of early life factors associated with college students’ alcohol use disorders could be an important starting point.
2.2 literature review
College students typically spend a tremendous amount of time with their family members, emphasizing the influence of family quality on any type of habit acquisitions. Evidence has shown family relationship quality is strongly correlated with early alcohol use (Kelly et al. 2011; Brody and Forehand 1993), and the effect is interactive with gender (Kelly et al. 2011). Since studying also comprises an important part of college life, it is important to evaluate how college life and alcohol use interact with each other. An 21 year follow-up of 3,478 Australian since they were child has found level of academic performance predicts their drinking problems, independently of a selected group of individual and family con-founders (Hayatbakhsh et al. 2011). College students start to build up their social networks. An increased exposure to social communications is reasonably expected among them, which might incur alcohol involvement. A survey has found typical social drinking contexts were associated with men’s average daily number of drinks and frequency of drunkenness, indicating social communications, interacted with gender, might have influence on college students’ alcohol usage(Senchak, Leonard, and Greene 1998).
2.3 Proposing hypothesis
According to the literature review, 4 potential early-life factors is identified to predict excessive alcohol usage among college students. They are a. family relationship quality (interactive with gender); b. school performance; c. social communication (interactive with gender). I herein proposed a 3-factor alcohol high-use model for college students and test it using a secondary data set collected for other purposes.
In the data set, variables including gender, quality of family relationships (“famrel”), number of school absences (“absences”), weekly study time (“studytime”) and frequency of going out with friends (“goout”) could be candidate indicators for the current model. The variable “gender” and “famrel”‘s relevance to the predictors are self-explanatory. School performance includes college students’ in-class and off-class performance, which could be reflected by variables “absences” and “studytime”, respectively. Variable “goout” captures the involvement of social activity, which is a good indicator to social communication. Note that base on the well-reported evidence introduced above, gender will not enter the model independently. Instead, it will comprise interaction terms with family relationship quality and social communication, respectively, and then enter the model.
3 Data exploration
3.1 The distribution of the chosen variables
wrap.lab <- c("in-class performance (absences, smaller is better)",
"family relationship quality (famrel)",
"social (goout)",
"off-class performance (studytime)")
names(wrap.lab) <- c("absences", "studytime", "famrel", "goout")
alc %>%
select(absences, studytime, famrel, goout) %>%
pivot_longer(everything(),
names_to = "variable",
values_to = "value") %>%
ggplot(aes(x = value))+
geom_bar(width =1, fill = "white", color = "black")+
facet_wrap(~variable, scales = "free",
labeller = labeller(variable = wrap.lab))+
scale_fill_brewer(palette = "Greys") +
theme(panel.grid.major = element_blank(),
panel.grid.minor = element_blank(),
panel.background = element_rect(fill = "lightgrey"))+
labs(title = "Distribution of the interested variables",
x = "Values of each variable",
y = "Frequency")
The distribution of in-class performance reflected by class absences is skewed to right with very long tail. This indicates that most students have full or almost full attendance of class, while a small number of students might be absent for quite a number of classes. Other three variables are semi-numeric ordinal obtained by item with Likert-marks, with labeling value ranging from 1 to 5. It is found most Finnish students tend to be very social, with more around 2/3 of them being at the high end the choices. For off-class performance reflected by study time, it is found that most Finnish students spend 2~10 hours a week studying. For family relationship quality, it is surprising to find only roughly 1/3 of the students having good or very good family relationship quality, and none of them believe they have excellent family relationship quality.
3.2 exploring the association between faimly relationship quality and alcohol high-use
The variable “famrel” in original data set elicited quality of family relationships (numeric: from 1 - very bad to 5 - excellent). In the current analysis, it is selected as a candidate predictor for the model to reflect the same idea–quality of family relationship.
3.2.1 numerically explore the association
alc %>% count(high_use, famrel)## # A tibble: 10 × 3
## high_use famrel n
## <lgl> <dbl> <int>
## 1 FALSE 1 6
## 2 FALSE 2 9
## 3 FALSE 3 39
## 4 FALSE 4 128
## 5 FALSE 5 77
## 6 TRUE 1 2
## 7 TRUE 2 9
## 8 TRUE 3 25
## 9 TRUE 4 52
## 10 TRUE 5 23It is found the absolute sample of participants with very bad (level 1) and/or bad (level 2) family quality is very small in number (n = 8). Caution should be taken about the potential large error.
3.2.2 graphically explore the association
#adapt the titles for each wrapped graph
sex.labs <- c("Female", "Male")
names(sex.labs) <- c("F", "M")
#draw the bar plot
p1 <- alc %>%
ggplot(aes(x = factor(famrel), fill = high_use)) +
geom_bar(position = "fill", color ="black") +
facet_wrap(~sex, #warp by sex
labeller = labeller(sex = sex.labs)) + #label each
labs(x = "Family relationship quality (larger is better)",
y = "Proportion of high-user",
title =
"Proportion of alcohol high-use by family relationship quality and sex")+
theme(legend.position = "bottom")+ #adapt the legend position
guides(fill=guide_legend(title = "Alcohol high-use"))+ #define legend title
scale_fill_discrete(labels = c("FALSE" = "Non-high-user",
"TRUE" = "high-user"))+ #define legend text
scale_fill_brewer(palette = "Greys") #define color theme
p1
The value of the variable “famrel” includes numbers from 1 - very bad to 5 - excellent. In the current study, I presume that the intervals between each consecutive pair of value is consistent, and hence see it as a numeric variable.
According to the bar plot of proportion, the hypothesis of using the current variable in model fitting is validated. It is found that with the increasing of family relationship quality, the proportion of alcohol high-use decreases, except for female from a very bad (level 1) relationship family, which had a proportion of high-users at zero. However, this low proportion suffers from a risk of error due to the small sample in the level (n = 8). The result should be interpreted with caution.
To facilitate understanding, the variable’s name will be changed to family.quality according to the hypothesis.
3.2.3 re-code the variable of family relationship quality
alc <- alc %>%
mutate(family.quality = famrel) #create a new variable family.quality
#it has the same value with famrel3.3 exploring the association between school performance (absences) and alcohol high-use
The variable “studytime” in original data set captured participants’ weekly study time (numeric: 1 - <2 hours, 2 - 2 to 5 hours, 3 - 5 to 10 hours, or 4 - >10 hours). The variable “absences” in original data set captured participants’ number of school absences (numeric: from 0 to 93). It is presumed in the current analysis that they reflect off-class and in-class school performance, respectively, and hence they are selected as candidate predictors.
3.3.1 exploring the association between in-class performance and alcohol high-use
3.3.1.1 numerically explore the association
library(DT)
alc %>% group_by(high_use, sex) %>%
summarise(mean = mean(absences),
sd = sd(absences),
median = median(absences),
Q1 = quantile(absences, prob = 0.25),
Q3 = quantile(absences, prob = 0.75),
sampleSize = n()) %>%
datatable() %>%
formatRound(columns = c(3:4), digits = 2)From the table, it is found that both the frequency and median(Q1,Q3) of class absences differed greatly between alcohol high-users and non-high-users, indicating its validity in entering the model.
3.3.1.2 graphically explore the association
p2 <- alc %>%
ggplot(aes(x = high_use, y = absences, fill = high_use)) +
geom_boxplot() +
geom_jitter(width=0.25, alpha=0.5)+
facet_wrap(~sex, labeller = labeller(sex = sex.labs)) +
scale_fill_brewer(palette = "Blues")+
labs(x = "Alcohol high-user",
y = "Freuqncy of class absences",
title =
"Frequency of class absences by alcohol high-use and gender")+
theme(legend.position = "none")+
scale_x_discrete(labels = c("FALSE" = "Non-high-user",
"TRUE" = "high-user"))
p2
The box plot showed similar information to the previous table. No noticeable difference in proportions of absences can be observed between genders, and hence their interaction would not be considered in fitting the model.
3.3.1.3 rename the variable
To facilitate understanding, the name of variable “absences” will be changed to in.class.performance according to the hypothesis of current study.
alc <- alc %>%
mutate(in.class.performance = absences)3.3.2 exploring the association between off-class performance (study time) and alcohol high-use
3.3.2.1 numerically explore the association
alc %>% count(high_use, studytime)## # A tibble: 8 × 3
## high_use studytime n
## <lgl> <dbl> <int>
## 1 FALSE 1 56
## 2 FALSE 2 128
## 3 FALSE 3 52
## 4 FALSE 4 23
## 5 TRUE 1 42
## 6 TRUE 2 57
## 7 TRUE 3 8
## 8 TRUE 4 4From the table, it is found the sample of participants with long and very long (level 4 and 5) study time in alcohol high-user group is very small in number (n = 12). Caution should be taken about the potential large error.
3.3.2.2 graphically explore the association
p3 <- alc %>%
ggplot(aes(x = factor(studytime), fill = high_use)) +
geom_bar(position = "fill", color = "black") +
facet_wrap(~sex,
labeller = labeller(sex = sex.labs)) +
labs(x = "Study time ranges (larger is longer)",
y = "Proportion of high-user",
title = "Proportion of alcohol high-use by study time ranges and sex")+
theme(legend.position = "bottom")+
guides(fill=guide_legend(title = "Alcohol high-use"))+
scale_fill_discrete(labels = c("FALSE" = "Non-high-user",
"TRUE" = "high-user"))+
scale_fill_brewer(palette = "Greys")
p3
The levels of study time ranges include 1 - <2 hours, 2 - 2 to 5 hours, 3 - 5 to 10 hours, or 4 - >10 hours. Their intervals are inconsistent, and hence it is not appropriate to enter a model as numeric variables, and it will be transformed into a categorical variable.
According to the bar plot of proportion, it is found that with the increasing of study time, the proportion of alcohol high-use decreases, indicating its validity in entering the model. However, male with long study time (level 3) is an exception, which had the lowest proportion of high-users across the levels. Notably, this low proportion suffers from a risk of error due to the small sample in the level. To address the risk of error, the levels of study time ranges will be re-coded as Long study(original level 3 + original level 4), Moderate study(original level 2) and Light study (original level 1). Besides, no noticeable difference in proportions of study length can be observed between genders, and hence their interaction would not be considered in fitting the model.
To facilitate understanding, the name of variable “studytime” will be changed to off.class.performance according to the hypothesis.
3.3.3 re-code the variable of study length
alc <- alc %>%
mutate(off.class.performance =
case_when(studytime == 3 |studytime == 4~"Long study",
studytime == 2~"Moderate study",
studytime == 1~"Light study") %>%
factor(levels = c("Light study", "Moderate study", "Long study")))3.4 exploring the association between social communication frequency and alcohol high-use
The variable “goout” in original data set captured participants’ frequency of going out with friends (numeric: from 1 - very low to 5 - very high). It is presumed in the current analysis that it reflects social involvement, and hence it is selected as a candidate predictor.
3.4.1 numerically explore the association
alc %>% count(high_use, goout)## # A tibble: 10 × 3
## high_use goout n
## <lgl> <dbl> <int>
## 1 FALSE 1 19
## 2 FALSE 2 82
## 3 FALSE 3 97
## 4 FALSE 4 40
## 5 FALSE 5 21
## 6 TRUE 1 3
## 7 TRUE 2 15
## 8 TRUE 3 23
## 9 TRUE 4 38
## 10 TRUE 5 32From the table, it is found the sample of participants having very low frequency of social communication (level 1) is small in number (n = 21). Caution should be taken about the potential large error.
3.4.2 graphically explore the association
p4 <- alc %>%
ggplot(aes(x = factor(goout), fill = high_use)) +
geom_bar(position = "fill", color = "black") +
facet_wrap(~sex,
labeller = labeller(sex = sex.labs)) +
labs(x = "Social Communication Frequency (larger is more frequent)",
y = "Proportion of alcohol high-users",
title =
"Proportion of social communication frequency ranges by alcohol high-use and sex")+
theme(legend.position = "bottom",
plot.title = element_text(size = 10))+
guides(fill=guide_legend(title = "Alcohol high-use"))+
scale_fill_discrete(labels = c("FALSE" = "Non-high-user",
"TRUE" = "high-user"))+
scale_fill_brewer(palette = "Greys")
p4
According to the bar plot, the proportion of alcohol high-users changed tremendously across different levels of social communication, indicating good validity of our model hypothesis about this variable. There is a clear borderline between social communication levels 1-3 and levels 4-5, though the difference is varied across genders. The levels are hence re-coded into two–Infrequent (original level 1-3) and Frequent (original level 4+5). Its interaction with sex will also be considered in fitting the model. This corresponds to the finding of previous evidence(Senchak, Leonard, and Greene 1998).
4 Model fitting
4.1 fitting base on the original hypothesis
fit1 <- glm(high_use~ family.quality:sex + social:sex + off.class.performance + in.class.performance, data = alc, family = "binomial")
summary(fit1)##
## Call:
## glm(formula = high_use ~ family.quality:sex + social:sex + off.class.performance +
## in.class.performance, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.8260 -0.6946 -0.4982 0.6392 2.3077
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.05927 0.59970 -0.099 0.92127
## off.class.performanceModerate study -0.44754 0.31133 -1.438 0.15058
## off.class.performanceLong study -1.03288 0.43290 -2.386 0.01704 *
## in.class.performance 0.06651 0.02294 2.899 0.00374 **
## family.quality:sexF -0.37948 0.15263 -2.486 0.01291 *
## family.quality:sexM -0.34314 0.14677 -2.338 0.01939 *
## sexF:socialFrequent 0.92783 0.38133 2.433 0.01497 *
## sexM:socialFrequent 2.50131 0.39566 6.322 2.58e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 452.04 on 369 degrees of freedom
## Residual deviance: 350.52 on 362 degrees of freedom
## AIC: 366.52
##
## Number of Fisher Scoring iterations: 4All of the hypothesized predictors have at least one level being significant in the model. Comparing to light study participants, moderate study participants is not significant in predicting alcohol high-use. Hence,this variable will be dichotomized into Light study and moderate to long study for better model performance and parsimony of levels. The reason why it is not dichotomized into long study and moderate to short study is because the sample of long study category is extremely small, risking introducing error in our model.
4.2 re-code variable with insignificant levels
alc <- alc %>%
mutate(off.class.performance =
case_when(off.class.performance == "Light study"~ "Light study",
TRUE~ "Moderate to long study") %>%
factor(levels = c("Light study",
"Moderate to long study")))4.3 fitting the model again
fit2 <- glm(high_use~ family.quality:sex + social:sex + off.class.performance + in.class.performance, data = alc, family = "binomial")
summary(fit2)##
## Call:
## glm(formula = high_use ~ family.quality:sex + social:sex + off.class.performance +
## in.class.performance, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.8509 -0.6950 -0.4982 0.6364 2.3410
##
## Coefficients:
## Estimate Std. Error z value
## (Intercept) -0.06960 0.59922 -0.116
## off.class.performanceModerate to long study -0.58019 0.30223 -1.920
## in.class.performance 0.07155 0.02275 3.145
## family.quality:sexF -0.40471 0.15192 -2.664
## family.quality:sexM -0.34358 0.14686 -2.339
## sexF:socialFrequent 1.01880 0.37639 2.707
## sexM:socialFrequent 2.51453 0.39586 6.352
## Pr(>|z|)
## (Intercept) 0.90753
## off.class.performanceModerate to long study 0.05489 .
## in.class.performance 0.00166 **
## family.quality:sexF 0.00772 **
## family.quality:sexM 0.01931 *
## sexF:socialFrequent 0.00679 **
## sexM:socialFrequent 2.12e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 452.04 on 369 degrees of freedom
## Residual deviance: 352.88 on 363 degrees of freedom
## AIC: 366.88
##
## Number of Fisher Scoring iterations: 4Now all of the hypothesized predictors are significant in predicting alcohol high-use, except for off-class performance, which has a a p value of 0.05489, being very close to 0.05. An increase in sample size would very possibly make it significant. I hence keep this predictor in the model. Consequently, fit2 will be our final model.
4.4 interpreting the model results
4.4.1 transforming the coeficients to ORs
OR <- coef(fit2) %>% exp()
CI <- confint(fit2) %>% exp()
ORCI <- cbind(OR,CI)
print(ORCI, digits = 2)## OR 2.5 % 97.5 %
## (Intercept) 0.93 0.29 3.02
## off.class.performanceModerate to long study 0.56 0.31 1.02
## in.class.performance 1.07 1.03 1.13
## family.quality:sexF 0.67 0.49 0.90
## family.quality:sexM 0.71 0.53 0.94
## sexF:socialFrequent 2.77 1.33 5.86
## sexM:socialFrequent 12.36 5.84 27.75Our hypothesis that a. family relationship quality (interactive with gender); b. school performance; c. social communication (interactive with gender) could be predictors for alcohol high-use among college students is justified. According to the final model, comparing to participants who study less than 5 hours per week, those who study more than 5 hours have on average 0.56 (95%CI: 0.31~1.02) times less odds to be an alcohol high-user (95%CI: 0.31~1.02). Participants who have one more time of absence from class will on average have 1.07 (95%CI: 1.03~1.13) times more odds being an alcohol high-user. These findings about the predictive effect of academic performance on alcohol use is consistent with previous evidence(Hayatbakhsh et al. 2011).For female college students, every one unit of family relationship quality increase would lead to 0.67 (95%CI: 0.49~9.90) times less odds being alcohol high-user. For male students, every one unit of family relationship quality increase would lead to 0.71 (95%CI: 0.53~0.94) times less odds being alcohol high-user. These indicate the predictive effects of family relationship on alcohol use are present and different across genders. This finding is consistent with previous evidence (Kelly et al. 2011). For female college students, comparing to students who do not have social involvement frequently, those who usually have social engagement have 2.77 (95%CI: 12.36~5.84) times more odds of being alcohol high-users. For male students, this effect is also present but the effect size goes as high as 12.36 times more odds of being alcohol high-users. These indicate the predictive effects of social engagement on alcohol use are present and tremendously different across genders. This finding is consistent with previous evidence(Senchak, Leonard, and Greene 1998).
4.4.2 exploring predictions
4.4.2.1 cross tabulation of predcition versus the actual values
prob <- predict(fit2, type = "response")
alc <- alc %>%
mutate(probability = prob)
alc <- alc %>%
mutate(prediction = probability>0.5)
high_use <- alc$high_use %>%
factor(level = c("TRUE", "FALSE"))
prediction <- alc$prediction %>%
factor(level = c("TRUE", "FALSE"))
accuracy.table <- table(high_use = high_use, prediction = prediction)%>%
addmargins
accuracy.table ## prediction
## high_use TRUE FALSE Sum
## TRUE 54 57 111
## FALSE 23 236 259
## Sum 77 293 370#generate proportion of predictive performance
table(high_use = high_use, prediction = prediction) %>%
prop.table %>%
addmargins %>%
print(digits = 2)## prediction
## high_use TRUE FALSE Sum
## TRUE 0.146 0.154 0.300
## FALSE 0.062 0.638 0.700
## Sum 0.208 0.792 1.000#generate a function that calculates some indicators for sensitivity and specificity
my.fun <- function(array){
TP <- array[1,1] #true positive
FN <- array[1,2] #false negative
FP <- array[2,1] #false positive
TN <- array[2,2] #true negative
PP <- array[3,1] #positive
PN <- array[3,2] #negative
P <- array[1,3] # positive
N <- array[2,3] # negative
PPV <- TP/PP #positive predictive value
FOR <- FN/PN #false omission rate
FDR <- FP/PP # false discovery rate
NPV <- TN/PN # negative predictive value
TPR <- TP/P #true positive rate
FPR <- FP/N #false positive rate
FNR <- FN/P #false negative rate
TNR <- TN/N #true negative rate
a <- paste("Positive predictive value is", round(PPV,2))
b <- paste("False omission rate is", round(FOR,2))
c <- paste("False discovery rate is", round(FDR,2))
d <- paste("Negative predictive value is", round(NPV,2))
e <- paste("True positive rate is", round(TPR,2))
f <- paste("False positive rate is", round(FPR,2))
g <- paste("False negative rate is", round(FNR,2))
h <- paste("True negative rate is", round(TNR,2))
output <- list(a,b,c,d,e,f,g,h)
return(output)
}
my.fun(accuracy.table)## [[1]]
## [1] "Positive predictive value is 0.7"
##
## [[2]]
## [1] "False omission rate is 0.19"
##
## [[3]]
## [1] "False discovery rate is 0.3"
##
## [[4]]
## [1] "Negative predictive value is 0.81"
##
## [[5]]
## [1] "True positive rate is 0.49"
##
## [[6]]
## [1] "False positive rate is 0.09"
##
## [[7]]
## [1] "False negative rate is 0.51"
##
## [[8]]
## [1] "True negative rate is 0.91"Among 259 participants who are not alcohol high-users, our model correctly predicts 236 (91%) of them (True negative rate). Among 111 participants who are alcohol high-users, our model correctly predicts 54 of them (49%) of them (True positive rate). In all, among the 370 predicts, 80(21.6%) were inaccurate.
4.4.2.2 scatter plot of the prediction versus the actual values
library(dplyr); library(ggplot2)
p5 <- alc %>%
ggplot(aes(x = probability,
y = high_use,
color = prediction,
shape =factor(probability>0.5))) +
geom_point(position = position_jitter(0.01),
alpha = 0.8, size =2)
p5 
4.4.2.3 comparing the model to the performance of random guess
random.guess <- runif(n= nrow(alc), min = 0, max = 1)
alc <- alc %>%
mutate(random.guess = random.guess)
alc <- alc %>%
mutate(prediction.guess = random.guess>0.5)
high_use = alc$high_use %>%
factor(levels = c("TRUE", "FALSE"))
prediction.guess = alc$prediction.guess %>%
factor(levels = c("TRUE", "FALSE"))
accuracy.tab.rand<- table(high_use = high_use, prediction = prediction.guess) %>%
addmargins()
accuracy.tab.rand ## prediction
## high_use TRUE FALSE Sum
## TRUE 64 47 111
## FALSE 123 136 259
## Sum 187 183 370my.fun(accuracy.tab.rand)## [[1]]
## [1] "Positive predictive value is 0.34"
##
## [[2]]
## [1] "False omission rate is 0.26"
##
## [[3]]
## [1] "False discovery rate is 0.66"
##
## [[4]]
## [1] "Negative predictive value is 0.74"
##
## [[5]]
## [1] "True positive rate is 0.58"
##
## [[6]]
## [1] "False positive rate is 0.47"
##
## [[7]]
## [1] "False negative rate is 0.42"
##
## [[8]]
## [1] "True negative rate is 0.53"Among 259 participants who are not alcohol high-users, random guess correctly guesses 126 (57%) of them. Among 111 participants who are alcohol high-users, random guess correctly guesses 55 of them (47%) of them. In all, among the 370 predicts, 181(49%) were inaccurate. Our model shows a tremendously better overall performance than random guess. However, its effect on correctly predicting the alcohol high-users is roughly equal to random guess, indicating the model is better applied in predicting who is a non-alcohol-high-user.
4.5 cross validation (Bonus)
4.5.1 define loss function
# define a loss function (average prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}4.5.2 compute prediction error base on traing data set
# compute the average number of wrong predictions in the (training) data
training.error.full <- loss_func(alc$high_use, alc$probability)
training.error.full## [1] 0.2162162The prediction error rate is 21.6%, outperforming the model in Exercise Set 3, which had about 26% error.
4.5.3 compute prediction error base on 10-fold cross validation
# 10-fold cross-validation
set.seed(16)
library(boot)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = fit2, K = 10)
cross.val.error.full <- cv$delta[1]
cross.val.error.full## [1] 0.2216216According to the result of 10 fold cross-validation, the model has an average error rate of 22.2%, a bit larger than the results from training model, but the error rate is still notably lower than the model in Exercise.
5 Observing the relationship between prediction error and number of predictors (Super Bonus)
5.1 preparation
library(utils)#install.packages("utils") library for generating all possible combinations of n elements
#pass the name of 4 predictors for our final model into an object
used.predictor <- c("family.quality:sex",
"social:sex",
"off.class.performance",
"in.class.performance")5.2 generating prediction error for all possible combinations of subsets of selected predictors
#define a list "mylist"
mylist <- list()
#define a matrix with 2 rows and 6 columns, "ct.error", which means
#cross-validation and training error
ct.error <- matrix(nrow=2, ncol = 6)
#start a loop that generate all possible combinations of the 4 used predictors,
#the combination could have 1-4 elements. Four each number of element, start a
#loop (i in 1:4); Within the loop, another loop is used to pass all the prediction
#error results from cross validation and training data set into a matrix. Each
#Matrix will have two rows saving results of cv and training data set, respectively.
#The number of columns will be dependent on how many combinations will be produced,
#with the maximum number being 6 (number of possible combinations of 2 predictors).
#Base on the number of i, 4 matrices will be generated, and saved in mylist.
for(i in 1:4){
combinations <- combn(used.predictor, i)
all.formula.text <- apply(combinations, 2,
function(x)paste("high_use~",
paste(x, collapse = "+")))
for(j in 1:length(all.formula.text)){
all.formula <- as.formula(all.formula.text[j])
model <- glm(all.formula, data = alc, family = "binomial")
cv <- cv.glm(data = alc, cost = loss_func, glmfit = model, K =10)
ct.error[1,j] <- cv$delta[1]
alc <- mutate(alc, probability = predict(model, type = "response"))
ct.error[2,j] <- loss_func(alc$high_use, alc$probability)
}
mylist[[i]] <- ct.error
ct.error <- matrix(nrow=2, ncol = 6)
}
#collapse the 4 matrices in mylist into 4 data frames.
for(w in 1:4){
assign(paste0("df",w), as.data.frame(mylist[[w]]))
}
#merge the 4 data frames into 1 by row.
all.error <- rbind(df1,df2,df3,df4) #name the data set as all.error
#add a new column in all.error, which reflects if the result of this row is
#from cross validation or training set
tag <- rep(c("pred_cv", "pred_training"), times = 4)
#add another new column in all.error, which reflects if the result of this row is
#base on 1, 2, 3 or 4 predictors.
predictor_number <- rep(c(1,2,3,4), each = 2)
all.error <- all.error %>%
mutate(tag = tag,
predictor_number = predictor_number)
#calculate the mean and sd for each row.
#note that the rows base on 4 predictor will not have sd, since there is only
#one combination.
all.error <- all.error %>%
mutate(mean = rowMeans(select(.,V1:V6), na.rm = T),
sd = apply(.[,1:6], 1, function(x)sd(x, na.rm=T)))
#check the all.error data set
all.error## V1 V2 V3 V4 V5 V6 tag
## 1 0.3135135 0.2135135 0.3000000 0.2918919 NA NA pred_cv
## 2 0.2972973 0.2135135 0.3000000 0.2891892 NA NA pred_training
## 3 0.2135135 0.3027027 0.2675676 0.2135135 0.2108108 0.2783784 pred_cv
## 4 0.2108108 0.3054054 0.2648649 0.2135135 0.2054054 0.2837838 pred_training
## 5 0.2162162 0.2108108 0.2648649 0.2135135 NA NA pred_cv
## 6 0.2135135 0.2081081 0.2621622 0.2135135 NA NA pred_training
## 7 0.2216216 NA NA NA NA NA pred_cv
## 8 0.2162162 NA NA NA NA NA pred_training
## predictor_number mean sd
## 1 1 0.2797297 0.04503604
## 2 1 0.2750000 0.04124759
## 3 2 0.2477477 0.04014825
## 4 2 0.2472973 0.04299761
## 5 3 0.2263514 0.02577033
## 6 3 0.2243243 0.02535360
## 7 4 0.2216216 NA
## 8 4 0.2162162 NA5.3 plotting the trends of training&validation prediction errors by different number of predictors
#plot all.error
#the error ribbon is 95% confidence interval
#4 predictors (the fitted model) do now have a error range because there is only one combination
all.error %>% ggplot(aes(x = factor(predictor_number), y = mean, group = tag), color = tag) +
geom_line(aes(color = tag))+
geom_point()+
geom_ribbon(aes(ymin = mean-1.96*sd/sqrt(rowSums(!is.na(select(all.error,V1:V6)))),
ymax = mean+1.96*sd/sqrt(rowSums(!is.na(select(all.error,V1:V6)))),
fill = tag), alpha =0.25,
position = position_dodge(0.05))+
guides(fill = guide_legend(title = "Training/Cross-validation", title.position = "top"),
color = guide_legend(title = "Training/Cross-validation", title.position = "top"))+
theme_bw()+
theme(legend.position = c(0.82,0.85), axis.text.x = element_text(size=12),
panel.grid.major = element_blank(),
panel.grid.minor = element_blank()) +
labs(x = "", y = "Prediction Error",
title = "Trends of training&validation prediction errors by different number of predictors")+
scale_x_discrete(labels = c("1 predictor", "2 predictors",
"3 predictors",
"4 predictors \n(fitted model)"))+
scale_fill_discrete(labels = c("pred_cv" = "error base on cross-validation",
"pred_training" = "error base on training set"))+
scale_color_discrete(labels = c("pred_cv" = "error base on cross-validation",
"pred_training" = "error base on training set"))+
annotate("text",
label =
"↓4-predictor model does \nnot have CI since it only \nhas one type of combination",
x = 3.9, y = 0.24, color = "red")+
geom_point(aes(x=4, y= 0.22), shape = 1, size = 10, color = "red")
According to the plot, the prediction error of our 4-predictor final model is low comparing to the mean prediction errors of the possible combinations of either of the 1-,2-, and 3- predictor models. However, this goodness is only statistically significant comparing to 1- predictor models.(falls with 95%ci of 2- and 3- predictor models). This might be due to the possible combinations of 4 predictors is very small in number, resulting in large error ranges.
Super Bonus: Continued
Inspired by the bonus task, where different <4 number of predictors’ influence on prediction error was observed, I started to get interested in how different number of random combinations of predictors added to the final model would affect the error. There are almost 30 variables that were not used in the final model. Twenty-three of them do not have direct relationship with the entered predictors, and hence they were selected to be a free predictor pool. One to 15 different predictors were randomly selected from the pool, each with 100 random repetitions (if all possible combinations of the number of predictors <100, then all possible combination will be used), resulting in 1423 models. The error rate base on training dataset and 10 fold cross validation were computed and plotted in a line chart. 95% confidence interval for each number of added predictors were also calculated and visualized.
The reason why only 15 maximum added predictors will be used instead of all 23 predictors is because the current sample size could not faithfully support model with more than 19 predictors, according to a rule of thumb that for each predictor used in model, a sample of 20 is required.
Preparing the predictor pool
# The predictors used in final 4-factor model
fixed.predictor <- c("family.quality:sex",
"social:sex",
"off.class.performance",
"in.class.performance")
# The variables not used in final model
not.used.predictor <- c("sex", "famsize", "studytime", "famrel", "Dalc",
"Walc", "G1", "G2", "G3", "alc_use", "high_use",
"family.quality", "social", "probability",
"prediction", "random.guess", "prediction.guess",
"goout")
#The set of free predictor pool
free.predictor<- setdiff(names(alc), fixed.predictor)
free.predictor<- setdiff(free.predictor, not.used.predictor)Building a loop that generates the result of 1423 models with 5~19 predictors (4 predictors in final model are fixed)
mylist <- list()
ct.error <- matrix(nrow=2, ncol = 100)
for(i in 1:15){
combinations <- combn(free.predictor, i)
if(choose(23,i)>100){
ss = 100
}else{
ss = choose(23,i)
}
for(j in 1:ss){
rn <- round(runif(1,min = 1, max = choose(23,i)), 0)
sample.comb <- combinations[,rn]
formula.text <- paste(
"high_use ~ family.quality:sex + social:sex + off.class.performance + in.class.performance+",
paste(sample.comb, collapse = "+"))
model <- glm(formula.text, data = alc, family = "binomial")
cv <- cv.glm(data = alc, cost = loss_func, glmfit = model, K =10)
ct.error[1,j] <- cv$delta[1]
alc <- mutate(alc, probability = predict(model, type = "response"))
ct.error[2,j] <- loss_func(alc$high_use, alc$probability)
}
mylist[[i]] <- ct.error
ct.error <- matrix(nrow=2, ncol = 100)
}Collapsing the results into different data frame and merge them
for(w in 1:15){
assign(paste0("df",w), as.data.frame(mylist[[w]]))
}
all.error <- rbind(df1,df2,df3,df4,df5,df6,df7,df8,df9,df10,df11,df12,df13,df14,df15)
tag <- rep(c("pred_cv", "pred_training"), times = 15)
#add another new column in all.error, which reflects if the result of this row is
#base on 1-15 predictors.
predictor_number <- rep(c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15), each = 2)
all.error <- all.error %>%
mutate(tag = tag,
predictor_number = predictor_number)
#calculate the mean and sd for each row.
all.error <- all.error %>%
mutate(mean = rowMeans(select(.,V1:V100), na.rm = T),
sd = apply(.[,1:100], 1, function(x)sd(x, na.rm=T)))Plotting
#plot all.error
#the error ribbon is 95% confidence interval
#4 predictors (the fitted model) do now have a error range because there is only one combination
all.error %>% ggplot(aes(x = factor(predictor_number), y = mean, group = tag)) +
geom_line()+
geom_point()+
geom_ribbon(aes(ymin =
mean-1.96*sd/sqrt(rowSums(!is.na(select(all.error,V1:V100)))),
ymax =
mean+1.96*sd/sqrt(rowSums(!is.na(select(all.error,V1:V100)))),
fill = tag), alpha =0.25,
position = position_dodge(0.05))+
guides(fill = guide_legend(title = "Training/Cross-validation",
title.position = "top"),
color = guide_legend(title = "Training/Cross-validation",
title.position = "top"))+
theme_bw()+
theme(legend.position = c(0.2,0.15),
axis.text.x = element_text(size=12),
panel.grid.major = element_blank(),
panel.grid.minor = element_blank()) +
labs(x = "Number of random predictors added to final model(4 predictors)",
y = "Prediction Error",
title="Trends of training&validation error rates of final model plus \ndifferent number of random predictors")+
scale_fill_discrete(labels = c("pred_cv" = "error base on cross-validation",
"pred_training" = "error base on training set"))+
scale_color_discrete(labels = c("pred_cv" = "error base on cross-validation",
"pred_training" = "error base on training set"))+
geom_line(aes(y = 0.22), color = "coral", size = 0.2, alpha = 0.8)+
geom_line(aes(y = 0.21), color = "cyan3", size = 0.2, alpha =1)+
annotate("text",
label = "↓Error rate of final model base on cross validation",
x = 11, y = 0.22, vjust = -0.5)+
annotate("text",
label = "↓Error rate of final model base on training dataset",
x = 11, y = 0.21, vjust = -0.5)
It is found in the plot that the prediction error rate of the 4-predictor final model by cross validation is always lower than the mean prediction error (and their lower ends of confidence interval) of the final model plus 1 to 15 randomly selected predictors, indicating the goodness of our final model.
It is also interesting to observe that the more predictors introduced to the model, the error rate by training data set keeps decreasing, indicating more predictors produce better models. However, the results of error rate by cross validation show an opposite effect, where the error rates generally increase with more predictors (though some flucutations are present). Put together, it can be infered that measuring model error rates using training data set itself would lead to over-estimation of the model goodness when more predictors enter the model.
Reference
Chapter 4: Clustering and classification
I do not come up with any better way to arrange everything than follow the pathway of assignment requirement. I will report the analysis with each main section being one of the requirement and each subsection being a component of that requirement, consecutively. I hope this would also make your rating process easier.
1 Assignment requirement #1
Quoted from the assignment:
“Create a new R Markdown file and save it as an empty file named ‘chapter4.Rmd’. Then include the file as a child file in your ‘index.Rmd’ file.”
1.1 Create a new R Markdown file and save it as an empty file named ‘chapter4.Rmd’
Done!
1.2 include the file as a child file in your ‘index.Rmd’ file.
Done!
2 Assignment requirement #2
Quoted from the assignment:
“Load the Boston data from the MASS package. Explore the structure and the dimensions of the data and describe the dataset briefly, assuming the reader has no previous knowledge of it. (0-1 points)”
2.1 Load the Boston data from the MASS package.
# access the MASS package
library(MASS)
# load the data
data("Boston")
# pass Boston to another object for easy typing
bos <- Boston2.2 Explore the structure and the dimensions of the data
library(tidyverse)
#explore structure
str(bos)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...#explore dimensions
dim(bos)## [1] 506 14#generate a codebook
# string is copy from dataset introduction
codebook <- data.frame(variable = "CRIM - per capita crime rate by town/ZN - proportion of residential land zoned for lots over 25,000 sq.ft./INDUS - proportion of non-retail business acres per town./CHAS - Charles River dummy variable (1 if tract bounds river; 0 otherwise)/NOX - nitric oxides concentration (parts per 10 million)/RM - average number of rooms per dwelling/AGE - proportion of owner-occupied units built prior to 1940/DIS - weighted distances to five Boston employment centres/RAD - index of accessibility to radial highways/TAX - full-value property-tax rate per $10,000/PTRATIO - pupil-teacher ratio by town/B - 1000(Bk-0.63)^2 where Bk is the proportion of blacks by town/LSTAT - % lower status of the population/MEDV - Median value of owner-occupied homes in $1000's")
codebook <- codebook %>%
separate_rows(variable, sep = "/") %>% # "/" is the delimiter for rows
separate(variable, sep = " - ", #" - " is the delimiter for variables
into = c("name", "description"), # names of sparated variables
remove = T) #remove old column
#check codebook
library(DT)
codebook %>% datatable (caption = "Tab 2.2 Codebook for Boston dataset")The data set has 506 observations of 14 variables. Variable CHAS is a dummy variable where 1 means the the place having tract that bounds river, and 0 means otherwise. It needs to be converted to a factor.
bos <- bos %>%
mutate(chas = chas %>%
factor() %>%
fct_recode(
"With tracts that bonds river" = "1", #old value 1 to new label
"Otherwise" = "0") # old value 0 to new label
)2.3 describe the dataset
Each of the 506 rows in the dataset describes a Boston suburb or town, and it has 14 columns with information such as average number of rooms per dwelling, pupil-teacher ratio, and per capita crime rate. The last row describes the median price of owner-occupied homes.
3 Assignment requirement #3
Quoted from the assignment:
“Show a graphical overview of the data and show summaries of the variables in the data. Describe and interpret the outputs, commenting on the distributions of the variables and the relationships between them. (0-2 points)”
3.1 Show a graphical overview of the data
library(GGally)
library(ggplot2)
#define a function that allows me to fine-tune the matrix
my.fun <- function(data, mapping, method = "lm",...){ #define arguments
p <- ggplot(data = data, mapping = mapping) + #pass arguments
geom_point(size = 0.3,
color = "blue",...) + #define points size and color
geom_smooth(size = 0.5,
color = "red",
method = method) #define line size and color; define lm regression
p #print the results
}
#the abbreviated variable names are not self-explanatory, set column and row
#names to be the variable labels for better reading
#this new object will be used in ggpairs function
names1 <- pull(codebook[1:7,], description) # extract row 1：7 of var description
names1 <- sapply(names1, #collapse the description into multiple lines
function(x) paste(strwrap(x, 35), # for better reading
collapse = "\n")) # "\n" calls for a new line
ggpairs(bos,
lower = list(
continuous = my.fun,
combo = wrap("facethist", bins = 20)),
col = 1:7,
columnLabels = names1)+#define column labels as the names I just set
labs(title = "Fig 3.1 Visualized relations of Boston dataset, variable #1~#7" )+
theme(plot.title = element_text(size = 20) ) 
Note that variable about crime rate is plagued with outliers.
#repeat what is done in the last chunk for variable 8~14
names2 <- pull(codebook[8:14,], description)
names2 <- sapply(names1, function(x) paste(strwrap(x, 35), collapse = "\n"))
ggpairs(bos,
lower = list(
continuous = my.fun),
col = 8:14,
columnLabels = names2,
)+
labs(title = "Fig 3.1 Visualized relations of Boston dataset, variable #8~#14" )+
theme(plot.title = element_text(size = 20) ) 
3.2 Show summaries of the variables in the data.
library(finalfit)
#summarize the continuous data
ff_glimpse(bos)$Continuous %>% datatable(caption = "Fig 3.2 Summary of Continuous data")# summarize the categorical data
ff_glimpse(bos)$Categorical %>% datatable(caption = "Fig 3.2 Summary of Categorical data")3.3 Describe and interpret the outputs, commenting on the distributions of the variables and the relationships between them.
3.3.1 interpreting continuous variables
There are 13 continuous variables in the dataset. The crime rate of the town was 0.3(0.1~3.7)%; the proportion of a town’s residential land zoned for lots over 25,000 sq.ft. was 0 (0~12.5)%; the proportion of non-retail business acres per town was 9.7(5.2~18.1)%; the nitric oxides concentration was 0.5(0.4~0.6) parts per 10 million; the average number of rooms per dwelling was 6.3±0.7 rooms; the proportion of owner-occupied units built prior to 1940 was 77.5(45.0~94.1)%; the weighted distances to five Boston employment centres was 3.2 (2.1~5.2) kilometers; the index of accessibility to radial highways was 5(4~24) units of accessibility; the full-value property-tax rate was $330(279~666) per $10,000; the pupil-teacher ratio by town was 19.1(17.4~20.2); the Black proportion of population after taking the formula of 1000(Bk-0.63)^2 was 391.4(375.4~396.2); the proportion of population that is lower status was 11.4(6.9~17.0)%; the median value of owner-occupied homes was $21.2(17~25)*1000.
3.3.2 interpreting categorical variable
35(6.9%) towns have tracts that bonds Charles River.
3.3.3 commenting on the relationships between variables
Except for the one binary variable about tract that bonds river, each variable in our data set shows a >0.3 and/or <-0.3 correlation with at least one of the other variables. Some of them have correlation as high as 0.9. All of the correlation coefficients are significant (p<0.001).
4 Assignment requirement #4
Quoted from the assignment:
“Standardize the dataset and print out summaries of the scaled data. How did the variables change? Create a categorical variable of the crime rate in the Boston dataset (from the scaled crime rate). Use the quantiles as the break points in the categorical variable. Drop the old crime rate variable from the dataset. Divide the dataset to train and test sets, so that 80% of the data belongs to the train set. (0-2 points)”
4.1 Standardize the dataset and print out summaries of the scaled data
library(MASS)
#binary variables with values as number will not influence the result of
#standardization and clustering, hence I will reload Boston without re-coding
#binary variable. This is for easiness of matrix multiplication in the following
#operations
bos <- Boston
bos.s <- as.data.frame(scale(bos))# bos.s means Boston Scaled4.2 How did the variables change?
ff_glimpse(bos.s)$Con %>% datatable(caption = "Fig 3.2 Summary of standardized data")All the variables after scaling had a mean of 0 and most of variables’ values ranged from -4 and 4, only except for variables crim (crime rate), which might be due to out-liers (corresponds to the finding from the correlation matrix).
4.3 Use the quantiles as the break points in the categorical variable and drop the old crime rate variable from the dataset.
#generate cutoff according to quantile
bins <- quantile(bos.s$crim)
#generate a categorical variable "crime" and re-code it
bos.s <- bos.s %>%
mutate(crime = crim %>%
cut(breaks = bins, include.lowest = TRUE) %>%
fct_recode("Low" = "[-0.419,-0.411]",
"MediumLow" = "(-0.411,-0.39]",
"MediumHigh" = "(-0.39,0.00739]",
"High" = "(0.00739,9.92]"))
#remove crim
bos.s <- bos.s %>% dplyr::select(-crim)4.4 Divide the dataset to train and test sets, so that 80% of the data belongs to the train set
set.seed(2022)
#generate an object containing the number of observations in bos dataset
n <- nrow(bos.s)
#generate an object "ind", which contains a random selected set of the indexing
#of bos dataset, and the number of indexing takes up 80% of number of observations
ind <- sample(1:n, size = n*0.8)
#generate train&test sets according to the random set of indexing number
train <- bos.s[ind,]
test <- bos.s[-ind,]5 Assignment requirement #5
Quoted from the assignment:
“Fit the linear discriminant analysis on the train set. Use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables. Draw the LDA (bi)plot (0-3 points)”
5.1 Fit the linear discriminant analysis on the train set (Use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables)
# fit an linear discriminant model on the train set, named as "lda.fit"
lda.fit <- lda(crime ~ ., data = train) 5.2 Draw the LDA (bi)plot
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(factor(train$crime))#plot the lda results
plot(lda.fit, dimen = 2,
pch = classes,
col = classes,
main = "Fig 5.2 Biplot for LDA for clustering crime rate")+
lda.arrows(lda.fit, myscale = 4)
## integer(0)Biplot based on LD1 and LD2 was generated, see fig 5.2. The most of four clusters separated poorly, except for the cluster “High”. Heavy overlap was observed between each pair of other cluster. Besides, Clusters High and MediumHigh also showed notable overlaps.
Based on arrows, varaibles lstat explained the most for cluster High. Contributions of variables to other clusters are not clear enough due to the heavy overlap.
6 Assignment requirement #6
Quoted from the assignment:
“Save the crime categories from the test set and then remove the categorical crime variable from the test dataset. Then predict the classes with the LDA model on the test data. Cross tabulate the results with the crime categories from the test set. Comment on the results. (0-3 points)”
6.1 Save the crime categories from the test set and then remove the categorical crime variable from the test dataset
#save crime into an object
classes.test <- test$crime
#remove crime
test$crime <- NULL6.2 predict the classes with the LDA model on the test data
predicted.test <- predict(lda.fit, test)6.3 Cross tabulate the results with the crime categories from the test set.
#generate a table that evaluate the accuracy of model, and pass the table into
#an object named "accuracy.tab"
accuracy.tab <- table(correct = classes.test, predicted = predicted.test$class )
#show the accuracy table
accuracy.tab## predicted
## correct Low MediumLow MediumHigh High
## Low 4 7 0 0
## MediumLow 10 16 4 0
## MediumHigh 2 11 17 1
## High 0 0 0 30#ask R to identify the correct predictions and add them up
correct.n = 0 # the number of correct predictions starting at 0
for (i in 1:4){ #4 loops because we have 4 rows/columns
correct.c <- accuracy.tab[
which(rownames(accuracy.tab) == colnames(accuracy.tab)[i]),
i] # if a cell has same row and column names, pass its value into "correct.c"
correct.n = correct.c+ correct.n # update the value of correct prediction
} # by adding "correct.c"
# calculate the percent of correct predictions for test set
correct.n/(nrow(bos.s)*0.2) #denominator is the number of obs. in test set## [1] 0.66205537 Assignment requirement #7
Quoted from the assignment:
“Reload the Boston dataset and standardize the dataset (we did not do this in the Exercise Set, but you should scale the variables to get comparable distances). Calculate the distances between the observations. Run k-means algorithm on the dataset. Investigate what is the optimal number of clusters and run the algorithm again. Visualize the clusters (for example with the pairs() or ggpairs() functions, where the clusters are separated with colors) and interpret the results. (0-4 points)”
7.1 Reload the Boston dataset and standardize the dataset
#reload Boston
data("Boston")
bos <- Boston
#standardize the dataset
bos.s <- as.data.frame(scale(bos))7.2 Calculate the distances between the observations
dis_eu <- dist(bos.s)
summary(dis_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.39707.3 Run k-means algorithm on the dataset
bos.s.km <- kmeans(bos.s, centers = 4) 7.4 Investigate what is the optimal number of clusters
date()## [1] "Tue Nov 29 12:55:17 2022"set.seed(22) #22 is the date I carried out the analysis
# determine the number of clusters
k_max <- 10
# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(bos.s, k)$tot.withinss})# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')+
geom_line(aes(x = 3, color = "red")) +
annotate("text",
label =
"←Elbow effect happens here",
x = 4.8, y =3800, color = "red") +
labs(title = "Fig 7.4 Trends of within-cluster sum-of-square with increasing number of k",
x = "Number of k",
y = "With-cluster SS")+
theme(legend.position = "none",
panel.background = element_rect(color = "black"))
There is a huge reduction in variation with K =3, but after that the variation does not go down as quickly. I will use K = 3 and do the k-means clustering again.
7.5 run the algorithm again
km <- kmeans(bos.s, centers = 3)7.6 Visualize the clusters
#define a function that allows me to fine-tune the matrix
my.fun.km <- function(data, mapping,...){ #define arguments
p <- ggplot(data = data, mapping = mapping) + #pass arguments
geom_point(size = 0.3,
color = factor(km$cluster),
...) + #define points size and color
stat_ellipse(geom = "polygon", mapping = mapping, alpha = 0.5)
#calculate an ellipse layer that separate clusters
p #print the results
}
ggpairs(bos, mapping = aes(fill=factor(km$cluster)),
lower = list(
continuous = my.fun.km
),
col = 1:7,
title = "Fig 7.6.1 Correlation Matrix with clusters, variables 1 to 7")+
labs(caption = "Note that variables with at least one level fully covered by one cluster will not produce ellipse clustering")+
theme(plot.title = element_text(size = 20) )
ggpairs(bos, mapping = aes(fill=factor(km$cluster)),
lower = list(
continuous = my.fun.km
),
col = 8:14)+
labs(caption = "Note that variables with at least one level fully covered by one cluster will not produce ellipse clustering", title = "Fig 7.6.2 Correlation Matrix with clusters, variables 8 to 14" )+
theme(plot.title = element_text(size = 20) )
7.7 interpret the results
By observing the elbow plot that depicts the how size of within-cluster sum-of-square changes with number of k, an optimal number of k = 3 was determined. The subsequent results of k-means clustering was visualized in a correlation matrix with containing variables in dataset. It is observed that some of the variables have contributed tremendously to the clustering. For example, the variable black separates 1 cluster with the other 2 clusters nicely, see the 5th column of the picture above (Fig 7.6.2), x axis; and the variable age separates a different cluster with the other 2 clusters roughly, see the 7th row of the picture above (Fig 7.6.1), y axis. Some pairs of the variables have also played important role in clustering, for example, the combination of black and dist variables separate the 3 clusters roughly, see picture above (Fig 7.6.2, column 1nd, row 5th) or see the picture below (Fig 7.6.3). Due to the limitation of presenting more dimensions in a 2-dimension screen, I am not able to dig into the clustering effect of more variables combined. Fortunately, k-means clustering has done that for me, mathematically.
bos %>% ggplot(aes(x = dis, y = black, color = factor(km$cluster))) +
geom_point() +
geom_abline(intercept = 480, slope = -25)+
geom_abline(intercept = 400, slope = -80) +
stat_ellipse(geom = "polygon",
aes(fill = km$cluster),
alpha = 0.25)+
theme(legend.position = "none",
panel.grid.major = element_blank(),
panel.grid.minor = element_blank())+
labs(x = "Weighted distances to five Boston employment centres",
y = "Proportion of blacks by town",
title = "Fig 7.7 the clustering effect of variables black and dis combined")
8 Assignment requirement #2
Quoted from the assignment:
“Bonus: Perform k-means on the original Boston data with some reasonable number of clusters (> 2). Remember to standardize the dataset. Then perform LDA using the clusters as target classes. Include all the variables in the Boston data in the LDA model. Visualize the results with a biplot (include arrows representing the relationships of the original variables to the LDA solution). Interpret the results. Which variables are the most influential linear separators for the clusters? (0-2 points to compensate any loss of points from the above exercises)“
8.1 Perform k-means on the original Boston data with some reasonable number of clusters (> 2)(Remember to standardize the dataset)
# k = 3 is the optimal clusters found
km <- kmeans(scale(Boston), centers = 3)8.2 Perform LDA using the clusters as target classes.
#reload and standardize data
bos.s <- as.data.frame(scale(Boston))
#save the clusters identified by k-means clustering as a column in the data set
bos.s$km.cluster <- km$cluster
lda.km <- lda(km.cluster ~ ., data = bos.s) 8.3 Visualize the results with a biplot (include arrows representing the relationships of the original variables to the LDA solution)
# target classes as numeric
classes <- as.numeric(factor(bos.s$km.cluster))
#plot the lda results as biplot
plot(lda.km, dimen = 2,
pch = classes,
col = classes,
main = "Fig. 8.3 Biplot for separating clusters identified by K means distance")+
lda.arrows(lda.km, myscale = 4)
## integer(0)8.4 Interpret the results
Biplot based on LD1 and LD2 was generated, see fig 8.3. The three clusters separated very clearly and some overlap observed between cluster 1 and cluster 3, and between cluster 2 and cluster 3. Cluster 1 and cluster 2 are perfectly separated.
Based on arrows, variables rm, dis and crim explained more for cluster 1; variables indus, rad, tax and nox explained more for cluster 2; and variables black, chas and ptratio explained more for clusters 3. Other variables’ role in clustering was much weaker.
9 Assignment requirement #9
Quoted from the assignment:
“Super-Bonus: Run the code below for the (scaled) train data that you used to fit the LDA. The code creates a matrix product, which is a projection of the data points. Install and access the plotly package. Create a 3D plot of the columns of the matrix product using the given code. Adjust the code: add argument color as a argument in the plot_ly() function. Set the color to be the crime classes of the train set. Draw another 3D plot where the color is defined by the clusters of the k-means. How do the plots differ? Are there any similarities? (0-3 points to compensate any loss of points from the above exercises)”
9.1 Run the code below for the (scaled) train data that you used to fit the LDA. The code creates a matrix product, which is a projection of the data points.
#reload the data
bos.s <- as.data.frame(scale(Boston))
#generate cutoff according to quantile
bins <- quantile(bos.s$crim)
#generate a categorical variable "crime" and re-code it
bos.s <- bos.s %>%
mutate(crime = crim %>%
cut(breaks = bins, include.lowest = TRUE) %>%
fct_recode("Low" = "[-0.419,-0.411]",
"MediumLow" = "(-0.411,-0.39]",
"MediumHigh" = "(-0.39,0.00739]",
"High" = "(0.00739,9.92]"))
#remove crim
bos.s <- bos.s %>% dplyr::select(-crim)
set.seed(2022)
#generate an object containing the number of observations
n <- nrow(bos.s)
#generate a random set of indexing number with n = 80% of the obs.
ind <- sample(1:n, size = n*0.8)
#generate the train set and test
train <- bos.s[ind,]
test <- bos.s[-ind,]9.2 Install and access the plotly package. Create a 3D plot of the columns of the matrix product using the given code. Adjust the code: add argument color as a argument in the plot_ly() function. Set the color to be the crime classes of the train set.
#select predictors for train set, with outcome variable removed
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication, saving the resulting matrix into matrix_product
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
#turn matrix into data frame
matrix_product <- as.data.frame(matrix_product)
#plot the 3D plot with LD1, LD2 and LD3
library(plotly)
p1 <- plot_ly(x = matrix_product$LD1,
y = matrix_product$LD2,
z = matrix_product$LD3,
type= 'scatter3d',
mode='markers',
color = train$crime, #Set the color to be the crime classes of the train set.
size = 2)
p19.3 Draw another 3D plot where the color is defined by the clusters of the k-means.
#select predictors for train set, with outcome variable removed
model_predictors <- dplyr::select(train, -crime)
#get the clusters of k-means for the train set
train.km <- kmeans(model_predictors, centers = 3)
p2 <- plot_ly(x = matrix_product$LD1,
y = matrix_product$LD2,
z = matrix_product$LD3,
type= 'scatter3d',
mode='markers',
color = factor(train.km$cluster), #color defined by clusters of the k-means
size = 1.5)
p29.4 How do the plots differ? Are there any similarities?
The LDA was trained according to a mathematical category of crime rates (quantiles), which has 4 categories. While k = 3 was adopted for the k-means clustering base on the size of within-cluster sum of square. Since LDA is a supervised technique, we know what are each categories represent, which are also labeled in the caption. K-means clustering is a unsupervised method and thus I do not know anything about the real-world representation of the 3 clusters identified before observing closely.
However, by observing the pictures together, it is interesting to find out that, cluster 3 from k-means nicely overlaps with High category from LDA. Also, cluster 2 from k-means roughly overlaps with Low and Medium low categories from LDA. As such, I will re-code categories from LDA according to this finding and see closely how well results from k-means and LDA are consistent.
9.4.1 Recode categories from LDA into High, Medium and Low (old Low + Medium Low)
train.crime3 <- train %>%
mutate(crime3 = crime %>%
fct_recode("Medium" = "MediumHigh" ,
"Low" = "MediumLow",
"High" = "High",
"Low" = "Low"))
km.cluster <- factor(train.km$cluster)
levels(km.cluster) <- c("Medium","Low","High")9.4.2 Check the accuracy table
accuracy.tab <- table(correct = train.crime3$crime3, kmean.pred = km.cluster)
accuracy.tab ## kmean.pred
## correct Medium Low High
## Low 10 187 15
## Medium 10 48 37
## High 5 0 929.4.3 Calculate the accuracy rate
# looping through the 3X3 matrix to add up the columns and rows with same name
correct.n = 0
for (i in 1:3){
correct.c <- accuracy.tab[which(rownames(accuracy.tab) == colnames(accuracy.tab)[i]), i]
correct.n = correct.c+ correct.n
}
# calculate the accuracy rate
correct.n/(nrow(bos.s)*0.8)## [1] 0.7139328It gets an accuracy rate of 71.3%, greatly outperforming the original LDA model, indicating k-mean cluster could be used as a cue for categorizing continuous variables.
6.4 Comment on the results
Overall, 66.2% of the predictions are correct, showing not quite satisfactory predicting effect of our linear discriminant analysis. Observe the result closely, it is found that the for high and medium high crime rate regions, the analysis did the best predictions, with 90% (47/52) of accuracy. For Low and medium low regions, the predictive effect of our analysis decreased tremendously. This might be the result of a. the violation of the assumption of multivariate normality (but evidence showed even when this is violated, LDA also exhibited good accuracy); b. large number of out-liers in the dependent variable before re-coding (LDA is sensitive to out-liers); c. The small size of category Low in test set; d. better categorization strategy for dependent variable needed (the current categorization is only base on quantiles, which is lack of more evidence-based foundation).